# Characteristic function and distribution of a random variable

This is exercise 4.3 in Bjork, Arbitrage Theory in Continous Time. $$X_t = \int^t_0 \sigma(s)dW_s$$ $$\sigma$$ is a deterministic function and $$W_t$$ is brownian motion.

I am asked to find the characteristic function of $$X_t$$ and thus showing that $$X_t$$ is normally distributed with mean zero and variance $$\int^t_0 \sigma^2(s)ds$$

I have found the characteristic function to be: $$E[e^{iuX_t}]= \exp \left[-u^2/2 \int^t_0 \sigma^2(s)ds \right]$$ How Can I conclude that $$X_t$$ is normally distributed then?

The characteristic function (chf) defines the distribution function in a unique correspondence. For $$X$$ Gaussian with mean $$0$$ and variance $$\sigma^2$$ the chf $$E[e^{i u X}]$$ which is given by $$e^{-\sigma^2 u^2/2}.$$ Thus if you identify the variance term then you are done. The characteristic function is the one of the Gaussian distribution. Thus the random variable $$X_t$$ is Gaussian.